如图,AB=AC,∠BAC=90°,BD⊥AE于D,CE⊥AE于E,且BD>CE. 求证:BD=EC+ED.

问题描述:

如图,AB=AC,∠BAC=90°,BD⊥AE于D,CE⊥AE于E,且BD>CE.
求证:BD=EC+ED.

证明:∵∠BAC=90°,CE⊥AE,BD⊥AE,∴∠ABD+∠BAD=90°,∠BAD+∠DAC=90°,∠ADB=∠AEC=90°.∴∠ABD=∠DAC.∵在△ABD和△CAE中∠ABD=∠EAC∠BDA=∠EAB=AC,∴△ABD≌△CAE(AAS).∴BD=AE,EC=AD.∵AE=A...